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seropon [69]
3 years ago
15

Your new motorcycle weighs 2450 N.

Physics
1 answer:
Whitepunk [10]3 years ago
5 0

Answer:

Explanation:

The negative only matters reallly if you are dealing with a 2d system. I could even define down as positive and up as negative. However, usually you are taught that down is negative. It really doesn't matter because a force is a force. It is only given a direction relative to another force or vector direction.

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What condition is necessary for the flow of heat? What analogous condition is necessary for the flow of charge?
cricket20 [7]
Temperature difference; voltage difference
6 0
4 years ago
A 250 GeV beam of protons is fired over a distance of 1 km. If the initial size of the wave packet is 1 mm, find its final size
Margarita [4]

Answer:

The final size is approximately equal to the initial size due to a very small relative increase of 1.055\times 10^{- 7} in its size

Solution:

As per the question:

The energy of the proton beam, E = 250 GeV =250\times 10^{9}\times 1.6\times 10^{- 19} = 4\times 10^{- 8} J

Distance covered by photon, d = 1 km = 1000 m

Mass of proton, m_{p} = 1.67\times 10^{- 27} kg

The initial size of the wave packet, \Delta t_{o} = 1 mm = 1\times 10^{- 3} m

Now,

This is relativistic in nature

The rest mass energy associated with the proton is given by:

E = m_{p}c^{2}

E = 1.67\times 10^{- 27}\times (3\times 10^{8})^{2} = 1.503\times 10^{- 10} J

This energy of proton is \simeq 250 GeV

Thus the speed of the proton, v\simeq c

Now, the time taken to cover 1 km = 1000 m of the distance:

T = \frac{1000}{v}

T = \frac{1000}{c} = \frac{1000}{3\times 10^{8}} = 3.34\times 10^{- 6} s

Now, in accordance to the dispersion factor;

\frac{\delta t_{o}}{\Delta t_{o}} = \frac{ht_{o}}{2\pi m_{p}\Delta t_{o}^{2}}

\frac{\delta t_{o}}{\Delta t_{o}} = \frac{6.626\times 10^{- 34}\times 3.34\times 10^{- 6}}{2\pi 1.67\times 10^{- 27}\times (10^{- 3})^{2} = 1.055\times 10^{- 7}

Thus the increase in wave packet's width is relatively quite small.

Hence, we can say that:

\Delta t_{o} = \Delta t

where

\Delta t = final width

3 0
3 years ago
What is the x-component of a vector with a magnitude of 115 km at an angle of 22°?
ruslelena [56]

The x-component of a vector are < 106.6, 43.07 >

Depending on the angle we are provided, the x-component of a vector can either be cos or sin. Cos always corresponds to the right triangle's side that contacts the specified angle.

If a vector v with magnitude ||v|| makes an angle θ with the positive x-axis then,

v = ||v|| cos θi + ||v|| sin θj

 =  < ||v|| cos θ , ||v|| sin θ >

Magnitude p = 115 km

Angle = 22°

p = ||p|| < cos θ, sin θ >

p = 115 < cos 22°, sin 22° >

p = 115 < 0.927, 0.3746 >

p = < 106.6, 43.07 >

Therefore,  the x-component of a vector are < 106.6, 43.07 >

Learn more about vectors here:

brainly.com/question/8043832

#SPJ1

4 0
2 years ago
A rock hanging from a string has a net force of zero acting on it ? true or false ?
sattari [20]

Answer:

true cuase it is true its not false

8 0
3 years ago
Un cuerpo de 60 kg se encuentra a una distancia de 3.5 m del otro cuerpo, de manera que entre ellos se produce una fuerza de 6.5
aev [14]

Answer:

1989.6Kg

Explanation:

The computation of the mass of the other body is given below:

As we know that

F = G × m1 × m2 ÷ r²

Here the G would have the constant value i.e. 6.67 × 10^-11Nm² / kg².  

Now

6.5 × 10^-7N = 6.67 × 10^-11Nm² / kg² × 60Kg × m2 / (3.5m) ²

m2 = (F × r²) / (G × m1)

m2 = (6.5 × 10^-7N × (3.5m) ²) ÷ (6.67 × 10^-11Nm² / kg² × 60Kg)

= 1989.6Kg

7 0
3 years ago
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